# Good Example Of Enter Full Name Here Essay

## Enter professor name here

Engineering Economics – Enter course code here

Engineering Geology

Problem 1:

The amount of initial investment at a rate of interest of 10% and after 3 years to get $4000. This can be used to tell investor’s their return value.

## Future value is give, interest rate and number of years.

P = F/ (1+i)n i.e. P = 4000/(1.03)5 = 3005. Interest rate is in percentage

## Using Time value of money equation for compound interest

This equation is used for the calculation if initial deposit for a return of 4000 after 3 years at 10% interest rate.

Rate of Interest and number of years are the variables.

We derive the formula from F = P(1+i)n. Placing the values gets us the initial deposit value for investment.

## Problem 2:

After not paying for the first 3 years, the amount owed now by the person taking the loan is to b calculated.

Rate of Interest s given. Loaned amount i.e. present value and the number of years loan was not given.

## Calculating the total interest after 3 years of non-payment

ie = (1+r/k)k – 1 i.e. ie = (1+0.08/4)4 – 1= 1.082 – 1 = .0824 = 8.24% (Using effective rate of interest formula

F = 500,000(1+0.0824)4 = 636,064 (calculating amount owed on 4th installment using time value of money)

F = 500,000(1+0.08)4 = 629,856 (calculating amount owed on 3th installment using time value of money if interest compounded annually)

F = 500,000 (1+0.08/365)1095 = 635,608 (Time value of money for daily interest i.e. 8% is annual interest. Divide by 365. Also, 3 years accumulation gives 365*3=1095)

## Interest rate and number of years are the variables.

Description of each step is in point 4.

Problem 3

Inflation rate= 8% and simple interest = 11%. Initial investment = 20000

Future value after years interest is F = 20000 * (1+5*.11) = 31000

The time value of money after 5 years of% inflation is F = F/(1+r)n = 31000/(1+.08)5 = 21098. As initial investment was 20,000, buying power retains.

Future value after ten years of 11% simple interest F = 20000*(1+10*.11) = 42000

Inflation after 10 years at 8% = 42000/(1+.08)10 = 19454. Buying power does not retain.

## Time value o money equation for future value and future cost of commodity after inflation are used.

Variables are inflation rate, number of years and interest rate.

Problem 4

The initial investment = 150000 Annual return after tax deduction = 22000/year R

Rate of return = 10% Maintenance cost at year 10 and salvage on year 20.

Return of Industry = 22000 *[1 – (1+0.1)-20]/0.1 = 187220 (the equation of Uniform series present worth is used to calculate total return after annual 22000 returns for 20 years)

Maintenance cost year 10 = 40000/(1+.1)10 = 15420 ( equation of single payment present worth factor is used to discount the maintenance cost to 10 years later which is the 20th year which is today)

Salvage value = 30000/(1+.1)20 = 4457 ( equation of single payment present worth is used to calculate the salvage value today i.e. 20 years. The stated value is at the time of purchase which was 20 years ago)

Present value = 187220-150000-15420+4457 = 9238

## The names of equation are present in step 3.

Uniform series present worth factor equation is used to calculate the total revenue generated at 20 years while using a compound interest rate of 10%. Single payment present worth factor equation is used to discount those payments to year 20, which is today.

## Interest rate and number of years are the variable.

Problem 5

Initial investments of steel and concrete buildings. Service life of both buildings and their yearly maintenance costs. Also, the yearly rent of a building is given.

## Using Equivalent Uniform annual series and capital recovery factor formula to calculate Steel building yearly cost at 15% interest

= 150000*[0.15/1-(1+0.15)-25] + 6000 = 29250

## Using Equivalent Uniform annual series and capital recovery factor formula to calculate Concrete building yearly cost at 15% interest

= 200000*[0.15/1-(1+0.15)-50] + 4000 = 34028

Cost of Rental per year = 32000

Using Equivalent Uniform annual series and capital recovery factor are used for calculation of uneven payments through a number of years to calculate the yearly payment required for the operation.

## Variables of the equation are number of years and interest rate.

Problem 6

Initial Investment = 71 million. Interest rates of 15% and 25%

## Using formula of Annual profit we use the following table

NPV (15%) = -71+80.65 = 9.69

NPV (25%) = -71 + 66.65 = -4.4

## Hence 15 % rate of return of feasible

The equations are used for calculation of yearly profit and Net present value respectively.

Problem 7

Total investment available 100000. Rate of return = 20%

Using NPV formula with net present value as zero for all three projects and Uniform series present worth factor formula for the accumulation of revenue’s current value

## Project A

0 = -50000 + 16600*[ROR at 9 years]

[ROR at 9 years] = 50000/16600 = 3.012

[1-(1+i)-9]/i = 3.012 i.e. solving it gives i=30%

## Project B

0 = -50000 + 15000*[ROR at 8 years]

[ROR at 8 years] = 50000/15000 = 3.33

[1-(1+i)-8]/i = 3.33 i.e. solving it gives i=25%

## Project C

0 = -100000 + 30000*[ROR at 6 years]

[ROR at 6 years] = 3.33

[1-(1+i)-6]/i = 3.33 i.e. solving it gives i=20%

## The best investment option is A & B as the all investment gets used and we get 55% return rate.

Problem 8

Plant A income = 800000/year

Plant B income = 840000/year

Total net income from upgraded plant B = 1780000

Salvage value of plant A = 120000

Expansion cost of Plant B = 1100000

Interest Rate = 11%

(a) Disassembling Plant A = 120000

Expanding plant B = -1100000

First three years will see 840000 * (1.11)3 = 1148810

Total cost before running of plants = 980000 – 1148810 = -168810 Unacceptable

(b) First 3 years Plant A = 800000* (1.11)3 = 1094104

Plant B = 1148810

Revenue = 2242914

Expansion cost + Salvage cost = -1100000 + 120000 = 980000

The total revenue generated is 2242914+980000 = 3222914 Acceptable

(c) Using Uniform series capital recovery factor too find even annual distributions

A = 1100000 * .11/[1-1.11-3] = 450134

As the revenue of the plant B itself is 1780000 which is annually (using A/P) = 247536 annually and 800000 plant A, it is acceptable.

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